PROFESSOR: Hi everybody.
Welcome back to recitation.
In lecture you did a bunch of examples of related rates
problems. So I have a couple more for you to do today.
So here we've got OK, so we've got air being blown into a
spherical balloon at a rate of 1,000 cubic
centimeters per second.
So the question is, how fast is the radius growing when the
radius is equal to 8 centimeters?
And then OK, so I've got a second question, which is, how
fast is the surface area growing at that same time?
So why don't you take a few minutes, work this one out for
yourself, come back, and we'll work it out together.
All right.
Welcome back.
So this question, like all related rates questions, has
the property that the calculus is typically very
straightforward, but that there's some geometric or
algebraic setup.
So in this case, it's straight up geometry.
So we have a balloon, we know it's a perfect sphere, we know
how fast the volume is changing.
So OK, but we need to know how fast the radius is changing.
So in order to do that we need to figure out a relationship
between the radius and the volume.
And then we can just do implicit differentiation like
we've been doing.
So for example, OK, so for a sphere--
the setup is not so bad for this first one--
so we know that for a sphere the volume is equal to 4/3 pi
times the radius cubed.
So that's the fundamental relationship between the
volume and radius of a sphere, and it's true for every sphere
everywhere in Euclidean space.
And we're given also, that the volume is changing at a
constant rate of 1,000 centimeters cubed per second.
So dv dt is just given to be 1,000.
You know, leave off the units at this point.
So the question is, what is dr dt?
That's what we're trying to figure out.
We're trying to figure out how fast the radius is changing at
the moment when the radius is equal to 8 centimers.
So how can we do that?
Well, this fundamental
relationship, it's an identity.
It always holds.
So that means we can differentiate it.
So if we take the derivative of this identity, well, V on
the left just becomes dv dt.
And on the right we want to do implicit differentation.
So here r is changing with respect to time.
r is a function of t.
Well, OK.
So 4/3 pi is a constant.
So that when we differentiate nothing happens.
4/3 pi.
And so now we differentiate r cubed with respect to t, so
that gives us 3r squared times dr dt.
So that's just the chain rule in action there.
And now what we want is this dr dt.
Right?
That's the thing that we're looking for, is how fast the
radius is growing.
So that's dr dt.
And we want it at the moment when r is equal to 8.
So when r is equal to 8 this implies that-- well, OK, so dv
dt is 1,000 always--
and it implies that--
OK, so it's equal to 1,000 is equal to 4/3 pi times 3 times
8 squared times dr dt.
So at this moment that we're interested in, we have this
equation to solve for dr dt, and this is a nice, simple
equation to solve.
You just divide through by everything on the right hand
side other than dr dt.
So this implies that dr dt is equal to-- well, OK, so I have
to divide 1,000 by all this stuff.
I think it works out to something like 125 over 32 pi.
All right.
So that's the exact value.
Maybe you're interested in sort of knowing about how
large this is.
So 32 pi is pretty close to 100, so this
is about 1.2 something.
So, all right.
So there we go.
So that answers the first question.
At that moment the radius is growing at a rate of 125 over
32 pi centimeters per second.
OK.
So now how about the second question that we've got here?
What about the surface area?
So again, we know how fast now the radius is changing and we
know how fast the volume is changing.
So in order to figure out how fast the surface area is
changing, we need something that relates the surface area
to either the volume or the radius.
Now the relationship between surface area and volume is
something that we could sort of work out if we had to, but
it's a lot easier to write down the surface area in terms
of the radius.
So let's do that.
So we have, I'm going to use the letter S to denote surface
area of a sphere.
So again, it's a general identity, you know, a
geometric fact that the surface area of a sphere is
equal to 4 pi times the radius squared.
And this is always true.
And now the thing that we want is the rate of change of the
surface area.
So the rate of change is the derivative.
So we want to compute the derivative here, ds dt.
So OK, so we just do it.
So ds dt is equal to- well, 4 pi hangs around.
And again, we differentiate r squared.
r is a function of t, so we have to use
the chain rule here.
So this is times 2r times dr dt.
So this is an identity.
So this is true always.
And now we want to know, at this particular moment in
time, when r is equal to 8, what is ds dt?
And in order figure that out, well OK, we just have to be
able to plug in for everything else.
So when r equals 8-- all right, well luckily, now if we
were just starting this problem from scratch here,
we'd have a problem.
Which is we wouldn't know what dr dt was.
But luckily, we've already figured it out right, so in
the first part of the problem.
So we know that when r is equal to 8, dr dt is equal to
125 over 32 pi.
Did I copy that right?
Yes, I did.
OK.
So OK, so in this case, the equation we have to solve is
just completely straightforward.
We just plug in the values and it's already solved for us.
So that's nice.
So that we get that ds dt at that moment is equal to--
well, it's 4 pi times 2 times 8 is the radius
times 125 over 32 pi.
Oh boy, and all right.
So we can work this out if we want, I guess.
That's 32 pi to cancel, so that's equal to 250.
And I guess the units there, better be centimeters squared
per second.
So at this moment in time the surface area is growing by 250
centimeters squared per second.
So that's all we had to do.
So we're all set.